"Solving Coefficient of x^n in Binomial Expansion

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Homework Help Overview

The discussion revolves around finding the coefficient of \( x^n \) in the binomial expansion of \( (1-x)^{-6} \). Participants are exploring the implications of the exponent being negative and the use of the generalized binomial series.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Some participants question whether a specific value for \( n \) is necessary to find the coefficient. Others suggest that the question may be asking for a general term's coefficient.
  • There is a discussion about using the generalized binomial series and how it applies to the given expression.
  • Participants are considering the implications of the formula for binomial series when the exponent is negative or fractional.
  • Questions arise regarding the simplification of the coefficient and whether the order of terms in the binomial expansion matters.

Discussion Status

The discussion is active with participants providing insights into the generalized binomial series and questioning the assumptions made in the original problem. Some guidance has been offered regarding the use of the series, but there is no explicit consensus on the approach or the simplification of the coefficient.

Contextual Notes

Participants note that the original problem lacks a specific value for \( n \), which may affect the ability to find a definitive coefficient. There is also a continuation of the question regarding finding coefficients in a different polynomial expression.

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Homework Statement



Find, in the simplest form, the coefficient of x^n in the binomial expansion of (1-x)^(-6).

Homework Equations





The Attempt at a Solution



i am not sure how to go about with this.
 
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are u sure you have to find a cofficient contain x^n?

because u should have a specific value for n, so that you could find the cofficient infront of it, or you should at least have which term you are looking for.
 
Willian93 said:
are u sure you have to find a cofficient contain x^n?

because u should have a specific value for n, so that you could find the cofficient infront of it, or you should at least have which term you are looking for.

Yes, that's the question. Maybe it's asking for the coefficient of for any term in the expression.
 
Since the exponent, -6, is not a positive integer, you will need to use the generalized binomial series:
(a+ b)^m= \sum_{k=0}^\infty \frac{m(m-1)\cdot\cdot\cdot(m-k+1)}{k!}a^kb^{m-k}

Here, of course, a= 1, b= -x, and m= -6 so this is

(1- x)^{-6}= \sum_{k=0}^\infty \frac{-6(-7)\cdot\cdot\cdot(-5-k)}{k!}(-1)^kx^{-6-k}

You, apparently, are asked for the coefficient when -6-k= n or when k= -6-n.
 
HallsofIvy said:
Since the exponent, -6, is not a positive integer, you will need to use the generalized binomial series:
(a+ b)^m= \sum_{k=0}^\infty \frac{m(m-1)\cdot\cdot\cdot(m-k+1)}{k!}a^kb^{m-k}

Here, of course, a= 1, b= -x, and m= -6 so this is

(1- x)^{-6}= \sum_{k=0}^\infty \frac{-6(-7)\cdot\cdot\cdot(-5-k)}{k!}(-1)^kx^{-6-k}

You, apparently, are asked for the coefficient when -6-k= n or when k= -6-n.

thanks

Is this answer the most simplified?

\frac{-6(-7)...(-11-n)}{(-6-n)!}

the general formula for binomial series for (a+b)^n is different when n is a positive integer and when n is a fractional or negative value?

(a+ b)^m= \sum_{k=0}^\infty \frac{m(m-1)\cdot\cdot\cdot(m-k+1)}{k!}a^kb^{m-k}

Does it matter if the powers(k and n-k) for a and b is swapped since a+b is commutative?

This is the continuation of the question:

Hence, find the coefficient of x^6 and x^7 in (1+2x+3x^2+4x^3+5x^4+6x^5+7x^6)^3
 
Last edited:
any further hints on this question?
 

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